`lim_(x rarr 0) ((2+x) sin (2+x) - 2 sin2)/(x) ` is equal to

Options

A. sin 2
B. cos 2
C. 1
D. 1 cos 2 + sin 2

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{(2+x) \sin(2+x) - 2 \sin(2)}{x} \), we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( x = 0 \) into the expression to see if we can directly evaluate the limit:
\[ (2+0) \sin(2+0) - 2 \sin(2) = 2 \sin(2) - 2 \sin(2) = 0 \]
The denominator also becomes \( 0 \) when \( x = 0 \). Thus, we have a \( \frac{0}{0} \) indeterminate form.

**Hint:** Check the form of the limit by substituting the value of \( x \) directly.

### Step 2: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

Here, let:
- \( f(x) = (2+x) \sin(2+x) - 2 \sin(2) \)
- \( g(x) = x \)

Now we need to differentiate \( f(x) \) and \( g(x) \).

**Hint:** Remember to differentiate both the numerator and the denominator.

### Step 3: Differentiate the numerator
Using the product rule and chain rule to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}[(2+x) \sin(2+x)] = \sin(2+x) + (2+x) \cos(2+x) \]
Now, we differentiate \( g(x) \):
\[ g'(x) = 1 \]

**Hint:** Use product and chain rules for differentiation.

### Step 4: Substitute into L'Hôpital's Rule
Now we substitute back into L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \left[\sin(2+x) + (2+x) \cos(2+x)\right] \]

**Hint:** Evaluate the limit by substituting \( x = 0 \) again.

### Step 5: Evaluate the limit
Now substituting \( x = 0 \):
\[ \sin(2+0) + (2+0) \cos(2+0) = \sin(2) + 2 \cos(2) \]

Thus, the limit is:
\[ \lim_{x \to 0} \frac{(2+x) \sin(2+x) - 2 \sin(2)}{x} = \sin(2) + 2 \cos(2) \]

### Final Answer
The limit is:
\[ \sin(2) + 2 \cos(2) \]